哈理工OJ 1971 Power of Two（拆分整数为2的次幂的和）

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题目链接：http://acm.hrbust.edu.cn/index.php?m=ProblemSet&a=showProblem&problem_id=1971

Power of Two
Time Limit: 2000 MS Memory Limit: 65536 K
Total Submit: 197(83 users) Total Accepted: 92(77 users) Rating: Special Judge: No
Description
You have been given a number, you should decomposite it into sum of power of two in descending order. For example:

240 = 128 + 64 + 32 + 16

and

115 = 64 + 32 + 16 + 2 + 1

Input
The first line contains an integer T(T ≤ 50), indicating the number of test cases. Each test case contains one number n(1 ≤ n ≤ 2^31-1) the number we are going to decomposite.

Output
For each test case, output “Case #i: ” followed by one line the conresponding result.

Sample Input
2
240
115

Sample Output
Case #1: 240 = 128 + 64 + 32 + 16
Case #2: 115 = 64 + 32 + 16 + 2 + 1

Source
“科林明伦杯”哈尔滨理工大学第三届ACM程序设计团队赛
Author
xiaodao

【思路分析】先把2的次幂都求出来，然后直接扫一遍就OK了。
【AC代码】

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<stack>#include<queue>using namespace std;#define ll long longll a[33];void init(){    for(ll i=0;i<=32;i++)    {        a[i]=(ll)pow(2,i);    }}ll re[35];int main(){    init();    int t,iCase=0;    scanf("%d",&t);    while(t--)    {        ll n,cnt=0;        scanf("%lld",&n);        ll x=n;        for(ll i=32;i>=0;i--)        {            if(n>=a[i])            {                n-=a[i];                re[cnt++]=a[i];            }        }        printf("Case #%d: %lld = %lld",++iCase,x,re[0]);        for(ll i=1;i<cnt;i++)        {            printf(" + %lld",re[i]);        }        printf("\n");    }    return 0;}

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